3
votes
1answer
43 views

Sum as an integral

Recently I have encountered weird notation that I don't see into. When I have some infinite sum $$\sum_{n=1}^{\infty}f(n)$$ I would rewrite it without thinking to the integral form like this ...
1
vote
0answers
38 views

Turning a summation into an integral

I have a summation of the form: $$y(x) = \sum\limits_{h=-L}^L\frac{A(h)\cdot R(h)^2}{((x-h)^2+R(h)^2)^{3/2}}$$ Where I wish to solve/optimise $R(h)$ (leaving $A(h) = const/h$) or $R(h)$ and $A(h)$ ...
0
votes
1answer
27 views

I need to show that this sequence is increasing and I'm almost there but I need help on last step.

Let $(1+\frac{1}{n})^n$ be a sequence and $f(x)=(1+\frac{1}{x})^x $ on $[1,inf)$. I need to show that f is non-decreasing by showing that $f'(x)\ge0$. So far I have: Let $g(x)=ln(f(x))$, where $ln$ ...
0
votes
2answers
29 views

Calculus power series

Hi could anyone help me to solve this. express the function $\int_x ^0 (\sin(t^2)\cdot \cos(t^2))$ as a power series. Because there is two trigo identies I do not know how to combine them to form a ...
5
votes
0answers
74 views

Beauty$\sum_{j,k=1}^\infty \frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5)$

Hi I am trying to calculate the infinite double sum $$ S:=\sum_{j,k=1}^\infty \frac{H_j(H_{k+1}-1)}{jk(k+1)(j+k)}=-\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5),\quad H_n:=\sum_{k=1}^n\frac{1}{k}\ \ ...
2
votes
1answer
45 views

What is this sequence of polynomials?

NovaDenizen says the polynomial sequence i wanted to know about has these two recurrence relations (1) $p_n(x+1) = \sum_{i=0}^{n} (x+1)^{n-i}p_i(x)$ (2) $p_{n+1}(x) = \sum_{i=1}^{x} ip_n(i)$ == i ...
2
votes
1answer
68 views

Hypergeometric Function simple identity

I must proove this property but I really have no idea of how to proove it: $${}_2F_1(a,b;c;z)=(1-z)^{-a}{}_2F_1(a,c-b;c,\frac{-z}{1-z}) $$ It seems its a 'simple' property, but I haven't been able to ...
1
vote
1answer
43 views

Prove $\ln\big(\frac{21}{10}\big) \leq \sum_{n=10}^{20} \frac{1}{10n}\leq \ln\big(\frac{20}{9}\big).$

Please help. I have been trying his question for the past 2 hours and cant seem to go anywhere with it. $$ \ln\big(\frac{21}{10}\big) \leq \sum_{n=10}^{20} \frac{1}{n}\leq \ln\big(\frac{20}{9}\big). ...
2
votes
3answers
57 views

Prove that $\lim_{n\to\infty} H_n/n = 0$ ($H_n$ is the $n$-th harmonic number) using certain techniques

I can't seem to use certain methods such as $\varepsilon$-N, L'Hôspital's Rule, Riemann Sums, Integral Test and Divergence Test Contrapositive or Euler's Integral Representation to prove that ...
2
votes
0answers
54 views

Is this Neumann series solution unique?

I have a Fredholm integral equation of the second kind given as $$f(x)=g(x)+\lambda\int_{-\infty}^\infty K(x,y)f(y)dy, $$ where $\lambda\in(0,1)$, the kernel $K(x,y)=\phi(x-y)$ is a Gaussian ...
1
vote
2answers
47 views

Not sure which test to use?

Trying to determine if the following series is convergent: $$\sum_{k = 1}^{\infty} {2^k ln(1+1/(3^k))}$$ I have no idea how to compute the integral so im not sure if I should use the integral test, ...
1
vote
1answer
43 views

Should I use the ratio test to determine convergence for $\sum_{k = 1}^{\infty}{1 \over k\left[1 + \ln^{2}\left(k\right)\right]}$?

I'm trying to determine whether this is convergent and I was wondering if using the ratio test would be the right way to do it? ${(k)(1+ln^2(k)) \over [k+1]\left[1 + \ln^{2}\left(k+1\right)\right]}$ ...
0
votes
1answer
49 views

Convergence of the infinite series $2^x\ln(1+1/3^x)$

The Q: determine whether the series converges or not $$\sum_{k=1}^\infty 2^k\ln(1+1/3^k) $$ So far I figured out that the function is positive and decreasing on [1,infinity). I decided to try ...
2
votes
3answers
38 views

standard Taylor series using substitution

Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$? Determine a range of validity for this series.
2
votes
1answer
70 views

Generalized sophomore's dream

It is well know that $$\int_0^1 x^{-x} dx=\sum_{n=1}^{\infty}n^{-n}.$$ I'm wondering if there is a way to characterize all the continous functions $f: \mathbb{R}^{+}\rightarrow \mathbb{R}$ such that ...
3
votes
3answers
103 views

Series Expansion Of An Integral.

I want to find the first 6 terms for the series expansion of this integral: $$\int x^x~dx$$ My idea was to let: $$x^x=e^{x\ln x}$$ From that we have: $$\int e^{x\ln x}~dx$$ The series expansion of ...
0
votes
2answers
68 views

Integral test for convergence?

Why does the series' terms have to be non-negative to use the integral test? Consider the series: $$\sum_{n = 1}^{\infty}\frac{n\cos n - \sin n}{n^2}$$ Even though it has negative terms, why can't ...
1
vote
0answers
30 views

Performance estimation of shellSort

I'm trying to make a performance estimation for shell-sort algorithm. And I fail in it. My formula: equals to where dz is outer while-loop, dy is middle for-loop, and dx is inner for-loop ...
1
vote
1answer
51 views

Prove that : $\lvert s_n - \frac \pi 4\rvert \le \frac 1 {2n+1}$, where $s_n = \sum^{n-1}_{j=0} \frac {(-1)^j} {2j+1}$

Prove (Leibniz' series): $|s_n - \frac \pi 4| \le \frac 1 {2n+1}, \forall n \in \mathbb N$ where $s_n = \sum^{n-1}_{j=0} \frac {(-1)^j} {2j+1} = 1 - \frac 1 3 + \frac 1 5$ ... To prove the result ...
2
votes
2answers
42 views

Is my divergence test correct?

This idea came to me while looking at the following graph of $f=\frac{1}{x}$: Now, the definite integral of $f$ from $1$ to $n$ is smaller than $f(1)+f(2)...f(n)$, from the graph above. But since ...
3
votes
2answers
49 views

Proving an integral using a series

If $f:(0,1]\rightarrow \textbf {R}$ is defined by $f(x)=2nx$ for $\frac{1}{n+1}\leq x \leq \frac 1n$ and $n$ is a natural number, assuming that $\sum_{k=1}^{k=\infty}1/k^2=\pi^2/6$, show that ...
3
votes
2answers
52 views

How to find this limit using integration?

What is the value of $$\lim_{n \to \infty}\frac{(\sum_{k=1}^{n} k^2 )*(\sum_{k=1}^{n} k^3 )}{(\sum_{k=1}^{n} k^6)}$$ I just know that it has to be done by converting it into an integral. I have no ...
0
votes
1answer
65 views

Find an expression for the area under the graph of f(x) as a limit?

f(X) = 2x/(x^2 +1), 1 <= x <= 3 Basically, I need to find an expression for the area under the graph within these intervals for the function as a limit. I understand the concept of the area ...
9
votes
2answers
154 views

Additional ways of defining real powers.

I am familiar with the following 3 way of defining real powers: Given $x,y\in\mathbb{R}$, such that $x\gt0$, we can define $$ x^y = \begin{cases} \sup_{\,q\in\mathbb{Q}}\{x^q :q\le y\} ...
1
vote
1answer
42 views

The sequence of indefinite integrals of a uniformly convergent sequence, converges uniformly

Question. Let $(g_n)$ be a sequence of Riemann-integrable functions, $$f_n(x)=\int_a^x {g_n(t)} \, \mathrm{d}t,$$ and $(g_n)$ uniformly convergent on $[a,b]$, then $(f_n)$ converges uniformly on ...
12
votes
4answers
226 views

Prove that $\displaystyle\int_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum_{n=1}^\infty \frac{1}{(n+a)^2}$

Prove that $$\int_{[0,1]} x^a(1-x)^{-1}\ln x \,dx = -\sum_{j=1}^\infty \frac{1}{(j+a)^2}$$ I know that we have a product of $x^a$, $\displaystyle\sum_{j=0}^\infty x^j$, and ...
7
votes
4answers
94 views

Proving that $\lim_{n \rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \sin(t^n) dt =0$

It is clear that $\lim_{n \rightarrow\infty} \int_{0}^{1} \sin(t^n) dt =0$. (which is not what is to be proved here) I don't know how to proceed with the remaining part of the integral ie $\lim_{n ...
8
votes
2answers
123 views

Find $\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$

I want to find the sum: $$\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$$ I start with finding the antiderivative of the integrant, which is: ...
0
votes
2answers
36 views

possible value of this series ? either convergent or in the borel sense?

given the series $$ f(x)= \sum_{n=0}^{\infty}\frac{(2x)^{n}}{n!}(-1)^{n}\frac{d^{n}}{dx^{n}}(\frac{1}{x-1})$$ how could i evaluate this for every x different from x=1 ?? thanks any hints? or if ...
0
votes
1answer
87 views

A new proof a non-linear Euler sum

According to Nielsen we have the following : If $$f(x)= \sum_{n\geq 0}a_n x^n $$ Then we have the following $$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt ...
18
votes
1answer
375 views

Prove that $\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots.$

Question. Let $f:[0,1]\to\mathbb R$ given by $$ f(x)=\left\{\,\,\, \begin{array}{ccc} \displaystyle{\left\lfloor{1\over x}\right\rfloor}^{-1}_{\hphantom{|_|}}&\text{if} & 0\lt x\le 1, \\ ...
12
votes
2answers
373 views

How to prove the closed form of an integral

How to prove the following identity: $${1\over 2\pi}\int_{0}^{2\pi}\ln(1-2r\cos x+r^{2})\,dx=2\ln r,\text{ where } r\gt 1.$$ I was asked to use the following result to prove it: ...
3
votes
3answers
148 views

Find the limit of $S_n=\sum_{i=1}^n \big\{ \cosh\big(\!\!\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$, as $n\to\infty$?

$S_n=\sum_{i=1}^n\big\{ \cosh\big(\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$ as $n\to\infty$ I stumbled on this question as an reading about Riemannian sums as in $$ \int_a^b f(x)\,dx =\lim_{x\to ...
0
votes
2answers
127 views

A question based on Lebesgue dominated convergence theorem.

Prove that $$\lim_{n\to \infty } \int_{-n}^{n}\left(1+\frac{x}{n}\right)^ne^{-x^2}dx$$ exists, and find its value. THX.
10
votes
8answers
537 views

Proving convergence of a sequence whose terms are integrals

How to prove the following sequence converges to $0.5$ ? $$a_n=\int_0^1{nx^{n-1}\over 1+x}dx$$ What I have tried: I calculated the integral $$a_n=1-n\left(-1\right)^n\left[\ln2-\sum_{i=1}^n ...
0
votes
2answers
84 views

Does $\sum_{m=1}^{\infty}f_{m}\left(x\right) $ converge for almost every $x$ in $X$?

Let {$f_{m}$} be a sequence of measureable real-valued functions in $\left(X,\mathrm{\mathcal{M}},\mu\right)$. Suppose ${\displaystyle \sum_{m=1}^{\infty}\left\{ {\displaystyle ...
22
votes
2answers
441 views

Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$

In the following thread I arrived at the following result $$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$ Defining $$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, ...
1
vote
1answer
54 views

Maclaurin Series of $\int_0^x \cos t^2\,dt$

Find the Maclaurin Series for $\int_{0}^{x}\cos t^2\,dt$. $$\cos(x) = \sum\frac{(-1)^n x^{2n}}{2n!}$$ I'm trying this: $$\cos^2 x = \sum\frac{(-1)^n x^{4n}}{(2n!)^2}$$ How would you solve this ...
3
votes
1answer
127 views

Parseval's Identity (Integral)

Calculate the integral: \begin{equation} \int_{-\pi}^{\pi}\left|\sum_{n=1}^{\infty}\frac{1}{2^{n}}e^{inx}\right|^{2}dx\end{equation} I'm familiar with Parseval's identity which states that for ...
3
votes
3answers
187 views

Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$

I want to derive an asymptotic expansion for the following Bessel function. I think I need to rewrite it in another form, from which I can integrate it by parts. I am interested in obtaining the ...
2
votes
0answers
53 views

Is it always possible to find a $t>0$, such that $\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|dx<C~~~?$

Is it always possible to find a $t>0$, such that $$\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$ where $C$ is independent of $n$. Here is my idea: We know that \begin{align} ...
1
vote
2answers
65 views

Prove that the series converges to the integral

Prove: $\int _0^{1}x^{-x}dx$ = $\sum_{n=1}^\infty\frac{1}{n^n} $ I thought of using: $x^{-x}$ = $e^{-x lnx}$ and then using : $e^{-xlnx}$ = $\sum_{n=1}^\infty\frac{(-xlnx)^n}{n!} $ but I'm stuck from ...
0
votes
1answer
76 views

Why does this integral equal $1$ when $n=1$?

As part of a boundary value problem, I did the following integral: $b_n = 2\int^1_0 (30 \sin(\pi x) - 100 (1-x)) \sin (n\pi x) dx$ next step: $b_n = 2\int^1_0 (30 \sin(\pi x)sin (n\pi x) - (100 ...
0
votes
2answers
723 views

Use Maclaurin Series to evaluate the definite integral correct to within an error $\lt 0.0001$

Definite integral: $$\int_0^{0.2} \dfrac{1}{1+x^5}\text{d}x$$ So I did series expansion of $$\sum_{n=0}^{\infty}\dfrac{(-1)^n\cdot x^{5n}}{5n+1}+C$$ and when I plug in $0.2$ that makes it ...
1
vote
1answer
162 views

Determine whether the series is convergent or divergent by expressing $S_n$ as a telescoping sum $\sum_{n=1}^{\infty}\frac{6}{n(n+3)}$

I have no idea where I'm going wrong or if I'm even doing this problem correctly. But here are my steps so far: $$\sum_{n=1}^{\infty}\frac{6}{n(n+3)}=S_n\sum_{i=1}^{n}\frac{6}{i(i+3)}$$ After ...
4
votes
1answer
102 views

Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$

I'm having trouble understanding infinite sequence and series as it relates to calculus, but I think I'm getting there. For the below problem: $$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$$ The solution ...
0
votes
1answer
99 views

Find an example of a positive function that its improper integral converge, but its series diverge

I need to find an example of a positive function f, and a constant a>0 such that the improper integral of f from 0 to infinity converge, but the series f(na) from 1 to infinity diverge.
0
votes
1answer
93 views

Integral and Summation Exchange

Why is it possible to do the following? $\int \space[\space \sum_{n=0}^\infty(-1)^n x^n \space]\space dx = \sum_{n=0}^\infty (-1)^n\int x^n dx$ I know that this is legal: $\int \sum_{n=0}^\infty ...
5
votes
3answers
207 views

Can any continuous function be represented as an infinite polynomial?

Can any continuous function be represented as an infinite polynomial? Motivation: the antiderivative $ \int^\ e^{-x^2}dx\ $ can be expressed as an infinite polynomial(write Taylor series for ...
3
votes
1answer
114 views

integration and limits of series expansion for coth

I have to calculate an integral that looks like this: $$ \int_0^{\infty} dx f(x) \coth(x) $$ where $f(x)$ is some function. To do this, I am using the following series expansion (that is taken ...